The fringe width in a $YDSE$ pattern is $2.4 \times 10^{-4} \, m$ when red light of wavelength $6400 \, \mathring{A}$ is used. How much will it change if blue light of wavelength $4000 \, \mathring{A}$ is used?

In a Young's double slit experiment,light of $500 \ nm$ is used to produce an interference pattern. When the distance between the slits is $0.05 \ mm$,the angular width (in degrees) of the fringes formed on the distant screen is close to $........^o$.

In $YDSE$ setup,light of wavelength $640 \, nm$ is used with $d = 0.8 \, mm$ and $D = 1 \, m$. If intensity at central maximum is $I_0$ and its position is $y = 0$,then:

$A$ beam of light consisting of two wavelengths $650 \, nm$ and $520 \, nm$ is used to illuminate the slits of a Young's double-slit experiment. The order of the bright fringe of the longer wavelength that coincides with a bright fringe of the shorter wavelength at the least distance from the central maximum is:

In Young's double slit experiment, the distance between the slits is $3 \text{ mm}$ and the slits are $2 \text{ m}$ away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength $480 \text{ nm}$ and $600 \text{ nm}$ respectively. The separation on the screen between the $5^{th}$ order bright fringes of the two interference patterns is

In Young's double-slit experiment,the maximum intensity is $I_0$. If one slit is closed,the new maximum intensity is: